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(D) The magnet is placed along $AB$ with its center at $O$. The length of the magnet is $2l = 10 	ext{ cm}$,so $l = 5 	ext{ cm} = 0.05 	ext{ m}$. T

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$10^{-4} 	ext{ T}$

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(D) The magnet is placed along $AB$ with its center at $O$. The length of the magnet is $2l = 10 	ext{ cm}$,so $l = 5 	ext{ cm} = 0.05 	ext{ m}$. The magnetic moment $M = 1 	ext{ Am}^2$.Point $C$ lies on the perpendicular bisector of the magnet,which corresponds to the equatorial position.The distance $r$ of point $C$ from the center $O$ of the magnet is the height of the equilateral triangle with side $a = 10 	ext{ cm} = 0.1 	ext{ m}$.$r = \sqrt{a^2 - (a/2)^2} = \sqrt{0.1^2 - 0.05^2} = \sqrt{0.01 - 0.0025} = \sqrt{0.0075} = \sqrt{75} 	imes 10^{-2} 	ext{ m} = 5\sqrt{3} 	imes 10^{-2} 	ext{ m}$.The magnetic field $B$ at the equatorial position is given by $B = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}}$.Since $r^2 + l^2 = (5\sqrt{3} 	imes 10^{-2})^2 + (5 	imes 10^{-2})^2 = (75 + 25) 	imes 10^{-4} = 100 	imes 10^{-4} = 10^{-2} 	ext{ m}^2$.Thus,$(r^2 + l^2)^{3/2} = (10^{-2})^{3/2} = 10^{-3} 	ext{ m}^3$.Substituting the values: $B = 10^{-7} 	imes \frac{1}{10^{-3}} = 10^{-4} 	ext{ T}$.

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$,find the magnetic induction at the point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

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